TECHNICAL NOTE 2.1 Travelling waves Overvoltage protection · 2019-09-24 · TECHNICAL NOTE OVERVOLTAGES The travelling speed is given to 1 v = -----L’ × C’ For vacuum and gases,

—TECHNIC AL NOTE 2 . 1

Travelling wavesOvervoltage protection

The TECHNICAL NOTES (TN) are intended to be used in conjunction with the

APPLICATION GUIDELINESOvervoltage protectionMetal-oxide surge arresters in medium-voltage systems.

Each TECHNICAL NOTE gives in a concentrated form additional and more detailed information about various topics of MO surge arrester and their application under normal and special service conditions.

First published August 2019

3TR AV EL L I N G WAV E S 3

1 Introduction

In direct current and alternating current circuits with frequencies of 16.7 Hz and 50 Hz or 60 Hz we have stationary or quasi stationary conditions, this means that at all points of an electrical con-ductor (e. g. a copper wire) the charges (electrons) drift homogeneously with an average speed of approximately 0.1 mm/s to 10 mm/s in the same direction. The current flow is initiated by an impulse (e. g. closing of a circuit breaker), which travels with the speed of light along the line.

Voltage and current impulses having a rise time shorter than the travelling time of an electromag-netic wave along the line, travel along the line in form of waves. This means that (disregarding damping) the current and voltage impulses travel along the line without changing their shape. Therefore, they are at another place a later point in time. Figure 1 shows as an example a voltage impulse travelling along a line with the speed v towards the line end. At the end of the line the voltage impulse is reflected, depending on the surge impedances ZE. Because we are dealing with overvoltage protection we concentrate on voltage impulses only.

2 Travelling waves

The theory of travelling waves is based on the wave equations and the solution of the telegraph equation. Only the capacitances and the induc-tances of a line are considered, disregarding the ohmic resistance per unit length and the conduc-tivity of the insulation, see Figure 2.

Current and voltage are connected to one another via the surge impedance ZL of the line (Ohms Law). The surge impedance ZL results from the inductance and capacitance per unit length of the line, disregarding the ohmic resis-tance per unit length and the conductance of the insulation. This means that the surge impedance ZL depends only on the geometry and the dielec-tric material (insulation) enclosing the conduc-tor. ZL is not depending on the place, this means the surge impedance ZL is the same along the entire line. In Table 1 typical values of Z are given for information.

L’ Z = ----- C’

L’ Inductance per unit length in H/kmC’ Capacitance per unit length in F/km

—Travelling waves

Any transient disturbance, such as a lightning stroke terminating on a phase conductor or the operation of a circuit breaker, can be analyzed by use of travelling waves. Travelling wave phenomena determine the protective distance of MO surge arresters.

—Figure 1: Voltage impulse travelling along a line

ZL Surge impedance of the lineZE Surge impedance connected at the line terminationu(t) Time dependent voltage impulsev Velocity / Travelling speedx Positive “forward” directiont1..t5 Timings

x

t₁

V

t₂ t₃

u(t)

t₄ t₅ZEZL

TECH N I C A L N OTE OV ER VO LTAG E S

The travelling speed is given to

1 v = ---------- L’ × C’

For vacuum and gases, we can assume µr = 1 and εr = 1, so that follows

1v = ---------- = c = 300 m/µs µ0 × ε0

for the travelling speed of surges in air.

µ0 = 1.256 × 10–6 Vs/Amε0 = 8.854 × 10–12 As/Vm

For insulating material different from air (or generally from gases) one can write

cv = ------ εr

As example, assuming for cables εr = 4 we come to a travelling speed of v = 150 m/µs in cables.

3 Reflection and transmission of travelling waves

As mentioned above, travelling waves travel with the same impulse wave shape along the line. This is expressed by the equation U(x,t) = Uf + Ur, which means that the voltage at each point of the line consists of a “forward” (in x-direction) travelling voltage wave Uf, and a “backward” travelling wave Ur.

When a voltage wave travelling on a line reaches a point of discontinuity, i.e a change in the surge impedance, part of the voltage is “reflected” backward and a part is “transmitted” forward. This means that voltage decreases, and voltage increases appear on the connection point of the two lines. The value of the voltage for each moment and for each place on the line is the sum of the respective present values of all voltage waves. Figures 3a and 3b show the principle situation. The surge impedance Z2 has half the value of Z1. For this reason, a negative reflection at the point of discontinuity occurs.

At both sides of the point of discontinuity the sum of the voltages must be the same. This means U1 = U2.

4

OHL HV transmission lines 420 kV Z ≈ 350 … 380 Ω

OHL HV sub-transmission lines 123 kV Z ≈ 450 Ω

OHL MV distribution lines 24 kV …36 kV Z ≈ 450 … 500 Ω

Cables Z ≈ 30 … 60 Ω

GIS and GIL Z ≈ 60 Ω

HV tower Z ≈ 200 Ω

Earthed shielding wire on HV tower Z ≈ 500 Ω

—Table 1: Typical values for surge impedance Z

Uf

Ur

Ufx

Line with open end

ZE

0

ZL UE

U = 2 × UE V

Z ∞E

C’u

i

L’ L’ L’ L’ L’ L’

x

C’ C’ C’ C’ C’ Z E

ZL

L1

L2

L3

T

—Figure 2: Electrical circuit of a loss less line

ZL Surge impedance of the lineZE Surge impedance connected at the line terminationu Voltage i Current x DirectionL’ Inductance per unit lengthC’ Capacitance per unit length

The reflected and transmitted parts of the volt-age impulse can be calculated with the reflection factor ru and the transmission factor bu. Both fac-tors depend on the surge impedances Z1 and Z2.

U2f 2 × Z2bu = ----- = -------- U1f Z1 + Z2

U1r Z2 – Z1ru = bu – 1 = ---- = -------- U1f Z2 + Z1

With the values for the impedances Z1 and Z2 in Figure 3 it follows for the transmission factor bu = 0.67 and for the reflection factor ru = –0.33. This means that a negative voltage of 0.33 × U1f is travelling “backwards”, and a voltage of U2f = 0.67 × U1f travels onwards. The voltage at the point of discontinuity is U1 = U2 = 0.67 × U1f.

4 Reflection at the end of a line

As explained above the travelling waves are changed at points of discontinuity and are re-flected at the end of a line. The voltage UE appear-ing at the end of the line depends on the imped-ance ZE terminating the line. In the following typical examples are given. In all the given exam-ples we assume a rectangular voltage impulse Uf travelling from a far-distance source in direction of the line end.

In Figure 4a the line has an open end, e. g. there is an open circuit breaker or an extreme high imped-ance. According the above equation for the reflec-tion factor we come to ru = 1. The voltage Uf will be completely reflected, and consequently the voltage UE becomes twice as high as the incoming voltage.

If the line is terminated according Figure 4b with a surge impedance ZE equal to the line surge imped-ance ZL there will be no reflection. The transmis-sion factor results to bu = 1. The reflection factor is ru = 0. It is the same as if the line is infinitely long.

TR AV EL L I N G WAV E S 5

—Figure 3: Reflection and transmission at points of discontinuity (principle depiction)

a: Before the voltage impulse reaches the point of discontinuity b: After the voltage impulse reached the point of discontinuity

Linienstärken für Grafiken

Linie oben: 1 pt, schwarz

Linie innen: 0.5 pt, grey03

Linie unten: 1 pt, grey03

Balkengrafiken:72 x 47 mm Aussenmass

Pfeil 7, 45 %

x

Z₁

U₁f

U₁=U₂=0

1Z₂ = –– Z₁ 2

L1

L2

L3

T






Pfeil 7, 45 %

x

Z₁

U₁f

U₁r

Uzf

U₁=U₂=U₁f–U₁r

1Z₂= –– Z₁ 2

L1

L2

L3

T

—Figure 4a: Line with an open end (ZE → ∞)

—Figure 4b: Line termi-nated with a matched impedance (ZE = ZL)






Pfeil 7, 45 %

Uf

Ur

Ufx

Line with open end

ZE

0

ZL UE

U = 2 × UfE

EZ → ∞

L1

L2

L3

T






Pfeil 7, 45 %Uf

0

Z = ZL E

U = UfE

L1

L2

L3

T

TECH N I C A L N OTE OV ER VO LTAG E S6

Figure 4c shows the extreme case of a short cir-cuit at the end of the line, ZE = 0. The reflection factor is ru = –1, and therefore the reflected volt-age is U1r = –Uf and consequently the voltage at the line end is UE = 0. This is quite logic, because at a short circuit the voltage must be zero.

In Figure 4d the line is terminated with a capaci-tance with the impedance Xc = 1 / (ω × C). Because of the very high frequency in the rise time of the rectangular voltage impulse the capacitor behaves in the first moment like a short circuit. Consequently, the voltage will be zero in the first moment. Later, when the frequency is zero (direct voltage), the impedance Xc becomes infinitely high and acts as an open end. Therefore, after a given time, depending on the time constant Tc, the voltage at the end of the line will be UE = 2 × Uf.

The opposite to case 4d is if the line is terminated with an inductance L with the impedance XL = ω × L, as shown in Figure 4e. In the first moment the inductance acts as an open end with the consequence that the incoming voltage will be doubled. Then the impedance will become zero, and the inductance acts as a short circuit. Consequently, the voltage will decay with the time constant TL to zero.

5 Calculation of voltage waves at points of discontinuity

Complex travelling wave problems are solved effectively by digital transient programs such as EMTP or ATP. For a principle understanding of the voltages occurring at points of discontinuity the lattice diagram suggested by Bewley can be used, see Figure 5. It shows the time dependent volt-age u(t) at the point of discontinuity, e. g. the connection point of different lines.

The voltage impulse u(t) shown in the lattice dia-gram above has a time duration of 3 µs and a peak value of U. If the voltage impulse reaches the point A the reflection and transmission of the voltage starts. After 1 µs on the time scale the value U is reached, U × ru is reflected and U × bu is transmitted forward, as shown in the lattice di-agram. After 2 µs on the time scale half the value of U reaches point A, and is reflected and trans-mitted, and so on.

With the same assumptions as in the example in Figure 3 (bu = 0.67 and ru = –0.33) the voltage u(t) at point A will rise to a value of 0.67 × U, the steepness of the voltage impulse will be decreased.

—Figure 4c: Line with a short circuit at the end (ZE = 0)

—Figure 4d: Line terminated with a capacitance (ZE = XC)

—Figure 4e: Line terminated with an inductance (ZE = XL)






Pfeil 7, 45 %Uf

Ur

0

U = 0

Short circuit

E

Z = 0E

L1

L2

L3

T






Pfeil 7, 45 %Uf

0

L1

L2

L3

T






Pfeil 7, 45 %Uf

0

L1

L2

L3

T

UE = 2 × Uf × (1 – e–t/Tc)

UE = 2 × Uf × e–t/TL

1ZE = Xc = ——— ω × C

ZE = XL = ω × L

TR AV EL L I N G WAV E S 7

6 Travelling current waves

Voltage and current waves are connected through the surge impedance Z. The current waves travel of course with the same speed as the voltage waves. The reflection and transmis-sion factors for current waves are different from the factors for voltage waves. The travelling wave theory for current waves is for instance import-ant for calculating the current through an MO surge arrester in the moment the arrester changes from the almost insulating state into the conducting state due to an overvoltage. This needs detailed analysis and is mentioned here for completeness only.

7 Summary

Travelling wave phenomena determine the protective distance of MO surge arresters.

—Voltage and current impulses with a rise time equal or shorter than the travel time along a line are at another place at a later point in time. Voltage and current are connected to one another via the surge impedance Z of the line.

Reflections and transmissions of the voltage occur at points of discontinuity, this means when two lines with different impedances are connected.

If a line end is terminated with a surge impedance of Z → ∞, a full positive reflection of the voltage occurs. In such a case the voltage is doubled and the steepness of the incoming voltage as well. In case of short circuit at the line end a full nega-tive voltage reflection occurs, and the voltage will be zero at this point.

Some typical cases of voltage reflections at a line end are shown, depending on the impedance Z at the line end.

Complex travelling wave problems are solved effectively by digital transient programs such as EMTP or ATP.

The principle of voltage reflection and transmis-sion at a point of discontinuity is explained with the lattice diagram suggested by Bewley.

—Figure 5: Lattice diagram (above) and the voltage u(t) at the point of discon-tinuity A (below).

x

0

3 µs

t

t

0

0

0.5 U × bu

2 µs

0.5 U × ru

0.5 UU × bu

1 µs

1 µs

U × ru

U

A

VU

U(t)

00

2 µs

Z₁

3 µs

1Z₂= –– Z₁ 2

1 µs

U

0.67 U

0.33 U

2 µs 3 µs

U(t)

x

0

3 µs

t

t

0

0

0.5 U × bu

2 µs

0.5 U × ru

0.5 UU × bu

1 µs

1 µs

U × ru

U

A

VU

U(t)

00

2 µs

Z₁

3 µs

1Z₂= –– Z₁ 2

1 µs

U

0.67 U

0.33 U

2 µs 3 µs

U(t)

1HC

013

88

71 E

N A

A

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abb.com/arrestersonline

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TECHNICAL NOTE 2.1 Travelling waves Overvoltage protection · 2019-09-24 · TECHNICAL NOTE OVERVOLTAGES The travelling speed is given to 1 v = -----L’ × C’ For vacuum and gases,